Mathematics

# Solve:$\int {{{\cos }^{ - 1}}x} \,dx$

##### SOLUTION

Consider the given integral.

$I=\int{{{\cos }^{-1}}xdx}$

$I=\int{1.{{\cos }^{-1}}xdx}$

We know that

$\int{uvdx=u\int{vdx}}-\int{\left( \dfrac{d}{dx}\left( u \right)\int{vdx} \right)}dx$

Therefore,

$I={{\cos }^{-1}}x\left( x \right)-\int{\left( -\dfrac{1}{\sqrt{1-{{x}^{2}}}} \right)\left( x \right)}dx$

$I=x{{\cos }^{-1}}x+\int{\dfrac{x}{\sqrt{1-{{x}^{2}}}}}dx$

Let $t=1-{{x}^{2}}$

$\dfrac{dt}{dx}=0-2x$

$-\dfrac{dt}{2}=xdx$

Therefore,

$I=x{{\cos }^{-1}}x-\dfrac{1}{2}\int{\dfrac{dt}{\sqrt{t}}}$

$I=x{{\cos }^{-1}}x-\dfrac{1}{2}\left( 2\sqrt{t} \right)+C$

$I=x{{\cos }^{-1}}x-\sqrt{t}+C$

On putting the value of $t$, we get

$I=x{{\cos }^{-1}}x-\sqrt{1-{{x}^{2}}}+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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